442 - Matrix Chain Multiplication
来源:互联网 发布:王珂假富豪 知乎 编辑:程序博客网 时间:2024/04/29 22:16
代码:
#include<cstdio>
#include<cstring>
#include<ctype.h>
struct stack_one{
int row, col;
}Matrix[30], Mul[50];
int main(){
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++){
char name[3];
int r, c;
scanf("%s%d%d", name, &r, &c);
Matrix[name[0] - 65].row = r;
Matrix[name[0] - 65].col = c;
}
char fuck[128];
while (scanf("%s", fuck) != EOF){
int len = strlen(fuck), j = 0, count = 0;
for (int i = 0; i < len; i++){
if (isalpha(fuck[i])){
Mul[j].row = Matrix[fuck[i] - 65].row;
Mul[j].col = Matrix[fuck[i] - 65].col;
j++;
}
if (fuck[i] == ')'){
if (Mul[j - 2].col != Mul[j - 1].row){
printf("error\n");
goto end;
}
count += Mul[j - 2].row * Mul[j - 1].row * Mul[j - 1].col;
Mul[j - 2].col = Mul[j - 1].col;
j--;
}
}
printf("%d\n", count);
end: ;
}
return 0;
}
思路:这道题也是关于栈的运用.(关于表达式的定义那,先是将表达式想复杂了,觉得用普通的栈无法完成.不知是题目出得不严谨,还是自己想多了).其实很简单,两个矩阵后面必为一个“)”.那么忽略“(”,只以“)”为标志,遇到“)”就将栈中的最后两个元素运算.依次类推!就能得出结果!
总结与反省:主要对题目内容和数据的抽象能力不够啊!特别是这里要注意忽略“(”!
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